Find the Max and Min values of f on the interval [1/2, 2]
for the equation:
f(x) = x - lnx
I am stuck on this, I tried to find the derivative but I can't figure it out. Any help would be greatly appreciated. Thanks.
The derivative of the function is just $\displaystyle f'(x)=1-\frac{1}{x}$. To find the max and mins, set this equal to zero and solve.
$\displaystyle 1-\frac{1}{x}=0$
These are the cricitcal points. The only point seems to be at $\displaystyle x=1$. This is in the given interval, so just find $\displaystyle (1,f(1))$. If look at the value of the derivative just after $\displaystyle x=1$, for example $\displaystyle f'(2)=\frac{1}{2}$, it's clear that the point $\displaystyle (1,f(1))$ is a minimum. Google the "first derivative test" if you're not familiar with this.
Edit: the point I made is actually incorrect if you read string6bean1977's reply. I should have used the "Closed Interval Method." However, the point $\displaystyle (1,f(1))$ is still a minimum.
I agree with adkinsjr, but would like to add that the values of x that satisfy f`(x)=0 must then be tested in f(x) to ensure continuity. The zeroes of f`(x) don't automatically make them "c" values. (critical) Additionally, undefined values, such as 0 in the denominator, may also qualify as critical numbers. Such as x=0, in this equation, makes it undefined. However, f(0) doesn't exist either (ln 0), so, in this case, 0 is NOT a "c" value.
Good luck!
BB